Posted on Sep 17, 2019

Daily Challenge #69 - Going to the Cinema

Today's challenge is to help John make an important financial decision!

John likes to go to the cinema, but he wants to find the most cost-effective way to go. He can buy a ticket for $15, or he can buy a membership card for $500. Every time he uses the membership card, the ticket will be 0.9 times the price he paid for the last one.

Ex. If John goes to the cinema three times:
A : 15 * 3 = 45
B : 500 + 15 * 0.90 + (15 * 0.90) * 0.90 + (15 * 0.90 * 0.90) * 0.90 ( = 536.5849999999999, no rounding for each ticket)

Create a function movie with three parameters: card (price of the card), ticket(normal price of a ticket), perc(fraction of what he paid for the previous ticket) and returns the first n such that ceil(price of B) < price of A

More examples:
movie(500, 15, 0.9)should return43
(with card the total price is 634, with tickets 645)

movie(100, 10, 0.95) should return 24
(with card the total price is 235, with tickets 240)

Good luck!

Today's challenge comes from CodeWars user g964. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (7)

TadPol • Sep 18 '19

Here is my try:

function movie(card, ticket,perc){ let i = 0 let ticketFullPrice = 0 let lowedTicket = ticket while(card>ticketFullPrice){ ticketFullPrice += ticket lowedTicket *= perc card += lowedTicket i++ } return i } movie(500,15,0.9) movie(100, 10, 0.95)

Mihail Malo • Sep 17 '19 • Edited

"first" lmao

const movie = (card, ticket, part) => { let i = 0 let a = 0 let b = card do { a += ticket b += ticket * part ** i i++ } while (Math.ceil(b) >= a) return i - 1 }

an "optimization" (no power)

const movie = (card, ticket, part) => { let i = 0 let a = 0 let b = card let cheap = ticket while (true) { a += ticket b += cheap cheap *= part if (Math.ceil(b) < a) return i i++ } }

"optimization II" (no Math.ceil, comparison with 0 instead of 1)

const movie = (card, ticket, part) => { let i = 0 let dif = -card - 1 let cheap = ticket while (true) { dif += ticket - cheap cheap *= part if (dif > 0) return i i++ } }

idiocy.

const movie = (card, ticket, part) => { let i = 1 let dif = -card - 1 let cheap = ticket while ((dif += ticket - (cheap *= part)) <= 0) i++ return i }

Yasser Beyer • Sep 17 '19

denied on first lol

Mihail Malo • Sep 17 '19

Check timestamp by hovering the date lol

Amin • Sep 17 '19 • Edited

Elm

module Movie exposing (movie) getReduction : Int -> Float -> Float getReduction times percentage = percentage ^ toFloat times profitableNumberOfMovies : Int -> Int -> Float -> Int -> Int profitableNumberOfMovies card ticket percentage times = let reduction : Float reduction = getReduction times percentage increasedCard : Int increasedCard = ceiling <| toFloat card + toFloat ticket * reduction increasedTicket : Int increasedTicket = ticket * times in if increasedCard < increasedTicket then times - 1 else profitableNumberOfMovies increasedCard ticket percentage <| times + 1 movie : Int -> Int -> Float -> Int movie card ticket percentage = profitableNumberOfMovies card ticket percentage 1

Tests

module MovieTest exposing (suite) import Expect import Movie exposing (movie) import Test exposing (Test) suite : Test suite = Test.describe "Movie" [ Test.test "It should return 43 for movie 500 15 0.9" <| \_ -> Expect.equal 43 <| movie 500 15 0.9 , Test.test "It should return 24 for movie 100 10 0.95" <| \_ -> Expect.equal 24 <| movie 100 10 0.95 ]

E. Choroba • Sep 17 '19 • Edited

#!/usr/bin/perl use warnings; use strict; sub movie { my ($card, $ticket, $perc) = @_; my $reduced = my $sum = $ticket * $perc; my $count = 0; $sum += ($reduced *= $perc) while $ticket * ++$count < $card + int $sum; return $count } use Test::More tests => 3; is movie(100, 200, 0.5), 1; is movie(500, 15, 0.9), 43; is movie(100, 10, 0.95), 24;

Yasser Beyer • Sep 17 '19 • Edited

JavaScript

const movie = (card, ticket, perc) => { let n = 1; let frac = ticket * perc; let a = ticket * n; let b = card + frac; while(a < Math.ceil(b)) { n++; frac = frac * perc; b = b + frac; a = ticket * n; } return n; }